For the record, multiplication by $i$ does not preserve $TS^{2n-1}$, because this is an odd-rank bundle. In fact, the space of complex tangencies to $S^{2n-1}\subset \mathbb{C}^n$ is the standard contact structure on $S^{2n-1}$.
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Marco GollaAug 27 '11 at 8:52

1 Answer
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The map is the restriction to $TS^{2n-1}$ of a quadratic map, so its differential is the restriction of the differential of that map. In coordinates $z^{\mu}$ on $\mathbb{C}^n$, we get coordinates $z^{\mu},\xi_{\mu}$ on $T^*\mathbb{C}^n$, and $z^{\mu},\xi^{\mu}$ on $T\mathbb{C}^n$. The function is $f=\left<\xi, \sqrt{-1} z\right>=-\sqrt{-1}\xi^{\mu} z^{\bar{\mu}} +\sqrt{-1} \xi^{\bar{\mu}} z^{\mu}$, so has differential $df=-\sqrt{-1}\left(\xi^{\mu} dz^{\bar{\mu}} + z^{\bar{\mu}} d\xi^{\mu}\right)+\sqrt{-1}\left(\xi^{\bar\mu} dz^{\mu} + z^{\mu} d\xi^{\bar\mu}\right)$. The equations on the sphere are $z^{\mu} z^{\bar{\mu}} = 1$ so the tangent bundle of the sphere is $z^{\mu} \xi^{\bar{\mu}} + \xi^{\mu} z^{\bar{\mu}} = 0$. Taking exterior derivative, you can simplify $df$ a little.